# Find the constant term in the expansion of?

((-4x^6)+(-2/x))^35

### 2 Answers

- θ βяιαη θLv 77 years agoFavorite Answer
Recall that by the Binomial Theorem:

(a + b)^n = ∑ [C(n, k) * a^(n - k) * n^k] (from k=0 to n).

So:

[-4x^6 + (-2/x)]^35 = ∑ [C(35, k) * (-4x^6)^(35 - k) * (-2/x)^k] (from k=0 to 35).

Then, using the laws of exponents to get x's exponent in terms of k:

∑ [C(35, k) * (-4x^6)^(35 - k) * (-2/x)^k] (from k=0 to 35)

= ∑ [C(35, k) * (-4)^(35 - k) * (x^6)^(35 - k) * (-2)^k * (1/x)^k] (from k=0 to 35)

= ∑ {C(35, k) * (-4)^(35 - k) * (-2)^k * x^[6(35 - k)] * x^(-k)] (from k=0 to 35)

= ∑ {C(35, k) * (-4)^(35 - k) * (-2)^k * x^[6(35 - k) - k]} (from k=0 to 35).

Now, the constant term's x exponent will be 0, so setting the exponent of x equal to 0 yields that the constant term occurs when:

6(35 - k) - k = 0 ==> k = 30.

Therefore, by evaluating C(35, k) * (-4)^(35 - k) * (-2)^k * x^[6(35 - k) - k] at k = 30, the constant term is:

C(35, 30) * (-4)^(35 - 30) * (-2)^30 * x^0 = C(35, 30) * 2^40.

I hope this helps!

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- Randy PLv 77 years ago
According to the Binomial Theorem the k-th term is C(35,k) (-4x^6)^k * (-2/x)^(35 - k)

The power of x is (x^6)^k * (x^-1)^(35 - k) = x^(6k) * x^(-35 + k) = x^(6k - 35 + k).

To be a constant term, this must be x^0. Set that expression equal to 0, solve for k, then plug k into the Binomial Expansion above.

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